Question

Write the definition of the following statement about sets?
{n^2 + n + 1: n∈ℕ } ⊆ {2n + 1 : n∈ℕ }

the answer provided is:
∀n∈ℕ ∃m∈ℕ (n^2 + n + 1 = 2m + 1}

The answer intuitively makes sense to me, but I don't know how it was derived

What I tried was
x∈{n^2 + n + 1: n∈ℕ } → x∈ {2n + 1 : n∈ℕ }

∃n∈ℕ (x = n^2 + n + 1} → ∃m∈ℕ (x = 2m + 1)

But i'm stuck here and can't get it to match the answer provided

Answer 1

this is essentially saying that every number of the n^2 + n + 1 can be written as 2m+1

or just prove n^2 + n + 1 is always odd.

Answer 2

take two cases ... n=2i or n=2i+1

Answer 3

well well that's better.

Answer 4

\[

n^2 + n+1=n(n+1) +1
\]
n(n+1) is always even hence n(n+1) +1 is odd

Answer 5

well I realized n^2 + n + 1 is always odd. What i'm confused about is sort of like why the answer was the way it is. I.e
∀n∈ℕ ∃m∈ℕ (n^2 + n + 1 = 2m + 1}

What steps that leads to the introducing the quantifiers ?

Answer 6

I prefer to write it out step by step that lead to
∀n∈ℕ ∃m∈ℕ (n^2 + n + 1 = 2m + 1}

Noticing that n^2 + n + 1 is odd somehow doesn't provide the connection between the two sets to me.

Answer 7

Let, \(m = 1 + 2 + + \dots +n =\frac{n(n+1)}{2}\)
then \( \forall n \in \Bbb N , \exists m \in \Bbb N | (n^2+n+1 = 2m+1) \)

Answer 8

huhm... This works fine